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6. In trapezoid $A B C D$, $A D / / B C, E F$ is the midline, the area ratio of quadrilateral $A E F D$ to quadrilateral $E B C F$ is $\frac{\sqrt{3}+1}{3-\sqrt{3}}$, and the area of $\triangle A B D$ is $\sqrt{3}$. Then the area of trapezoid $A B C D$ is | 6. 2 .
Let $A D=a, B C=b$, and the height be $2 h$. The area of trapezoid $A B C D$ is $S$. Then
$$
\begin{array}{l}
E F=\frac{a+b}{2}, \frac{a}{b}=\frac{S_{\triangle A B D}}{S_{\triangle B C D}}=\frac{\sqrt{3}}{S-\sqrt{3}} . \\
\text { Also, } \frac{\sqrt{3}+1}{3-\sqrt{3}}=\frac{S_{\text {quadrilateral } A E F D}}{S_... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,056 |
7. Given the equation $x^{2}+(2-k) x+1=0$ satisfies the condition $x>-1$, and $k x>0$ has only one real root. Then the range of values for $k$ is $\qquad$ . | 7. $k<0$ or $k=4$.
From the problem, we get $x_{1,2}=\frac{1}{2}\left(k-2 \pm \sqrt{k^{2}-4 k}\right)$.
Also, $\Delta=k^{2}-4 k \geqslant 0$, so $k \leqslant 0$ or $k \geqslant 4$.
By case analysis, we get $k<0$ or $k=4$. | k<0 \text{ or } k=4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,057 |
8. Given $\sqrt[3]{x^{2}}+\sqrt[3]{(35-x)^{2}}=13$. Then $\sqrt[3]{x}+\sqrt[3]{35-x}=$ $\qquad$ . | 8. 2,5 .
Let $\sqrt[3]{x}=m, \sqrt[3]{35-x}=n, a=m+n, b=m n$.
Then $m^{2}+n^{2}=13$, i.e.,
$$
a^{2}-2 b=13 \text {; }
$$
$m^{3}+n^{3}=x+(35-x)=35$, i.e.,
$$
a^{3}-3 a b=35 \text {. }
$$
From equations (1) and (2), eliminating $b$ gives
$$
a^{3}-39 a+70=0 \text {. }
$$
Solving yields $a=5,2,-7$ (-7 is discarded). | 2,5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,058 |
Example 6 The solution to the equation $6 x-3[x]+7=0$ is
$\qquad$
(21st "Hope Cup" Mathematics Invitational (Grade 8)) | Solve the equation $6 x-3[x]+7=0$ which can be rewritten as $[x]=\frac{6 x+7}{3}$.
Substitute it into $x-1<[x] \leqslant x$, to get $x-1<\frac{6 x+7}{3} \leqslant x$.
Solving this system of inequalities yields $-\frac{10}{3}<x \leqslant-\frac{7}{3}$.
Thus, $[x]=-4$ or -3.
Therefore, $\frac{6 x+7}{3}=-4$ or -3.
Solving ... | x=-\frac{19}{6} \text{ or } x=-\frac{8}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,059 |
9. Wang Qiang has four colors of small cylindrical rods, Table 1 lists the lengths of the rods of different colors.
Table 1
\begin{tabular}{|c|c|c|c|c|}
\hline Color & Green & Yellow-Red & Purple & Red \\
\hline Length & $3 \mathrm{~cm}$ & $4 \mathrm{~cm}$ & $8 \mathrm{~cm}$ & $9 \mathrm{~cm}$ \\
\hline
\end{tabular}
... | 9. 91.
From the problem, we have
$$
\begin{array}{l}
3(81+a)+4(81+b)+8(81+c)+9(81+d) \\
=2010,
\end{array}
$$
where $a, b, c, d$ are non-negative integers.
$$
\begin{array}{l}
\text { Simplifying, we get } 3(a+3 d)+4(b+2 c)=66 . \\
\text { Therefore, }(a+3 d, b+2 c) \\
=(4 i+2,15-3 i)(i=0,1, \cdots, 5) . \\
\text { W... | 91 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,060 |
10. In a competition with 20 singers, 9 judges respectively assign them ranks from 1 to 20. It is known that for each singer, the difference between any two ranks does not exceed 3. If the sum of the ranks each singer receives is arranged in an increasing sequence: $C_{1} \leqslant C_{2} \leqslant \cdots \leqslant C_{2... | 10. 24.
If 9 judges all give a singer the first place, then $C_{1}=9$.
If two singers both get the first place, then one of them gets no less than 5 first places, while the other 4 ranks are no higher than fourth place, so $C_{1} \leqslant 5 \times 1+4 \times 4=21$.
If three singers all get the first place, then the... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,061 |
1. Given an integer $n(n \geqslant 3)$, let $A_{1}, A_{2}, \cdots, A_{2 n}$ be pairwise distinct non-empty subsets of the set $\{1,2, \cdots, n\}$, and let $A_{2 n+1}=A_{1}$. Find
$$
\sum_{i=1}^{2 n} \frac{\left|A_{i} \cap A_{i+1}\right|}{\left|A_{i}\right| \cdot\left|A_{i+1}\right|}
$$
the maximum value.
(Liang Yingd... | 1. For any $i$, if $\left|A_{i} \cap A_{i+1}\right|=0$, then
$$
\frac{\left|A_{i} \cap A_{i+1}\right|}{\left|A_{i}\right| \cdot\left|A_{i+1}\right|}=0.
$$
Assume $\left|A_{i} \cap A_{i+1}\right| \geqslant 1$ below.
Since $A_{i} \cap A_{i+1} \subseteq A_{i}$ and $A_{i} \cap A_{i+1} \subseteq A_{i+1}$, we have $\left|A_... | n | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,062 |
2. As shown in Figure 1, in $\triangle ABC$, $AB = AC$, $D$ is the midpoint of side $BC$, and $E$ is a point outside $\triangle ABC$ such that $CE \perp AB$ and $BE = BD$. A line $MF \perp BE$ is drawn through the midpoint $M$ of segment $BE$, intersecting the minor arc $\overparen{AD}$ of the circumcircle of $\triangl... | 2. As shown in Figure 3, it is easy to see that $A D \perp B C$. Therefore, the circumcenter of $\triangle A B D$ is the midpoint $O$ of segment $A B$.
Extend $F M$ to intersect $\odot O$ at point $L$, connect $O E$, and draw $O H \perp F L, O K \perp A D$, intersecting $F L$ and $A D$ at points $H$ and $K$, respective... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,063 |
3. Prove that for each positive integer $n$, there exists a prime $p$ and an integer $m$ satisfying the following three conditions:
(1) $p \equiv 5(\bmod 6)$;
(2) $p \mid n$;
(3) $n \equiv m^{3}(\bmod p)$.
(Fu Yunhao) | 3. Proof 1 First prove: There are infinitely many primes that are congruent to 5 modulo 6.
If not, then there are only finitely many primes that are congruent to 5 modulo 6, let them be $p_{1}, p_{2}, \cdots, p_{r}$ in ascending order.
Consider the number $6 p_{1} p_{2} \cdots p_{r}-1$.
Since it is congruent to 5 modu... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,064 |
4. Let real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $\sum_{i=1}^{n} x_{i}^{2}=1$ $(n \geqslant 2)$. Prove:
$$
\begin{array}{l}
\sum_{k=1}^{n}\left(1-\frac{k}{\sum_{i=1}^{n} i x_{i}^{2}}\right)^{2} \frac{x_{k}^{2}}{k} \\
\leqslant\left(\frac{n-1}{n+1}\right)^{2} \sum_{k=1}^{n} \frac{x_{k}^{2}}{k},
\end{array}
$$
... | 4. Notice that
$$
\begin{array}{l}
\sum_{k=1}^{n} \frac{x_{k}^{2}}{k}\left(1-\frac{k}{\sum_{i=1}^{n} i x_{i}^{2}}\right)^{2} \\
=\sum_{k=1}^{n} \frac{x_{k}^{2}}{k}-\sum_{k=1}^{n} \frac{2 x_{k}^{2}}{\sum_{i=1}^{n} i x_{i}^{2}}+\sum_{k=1}^{n} \frac{k x_{k}^{2}}{\left(\sum_{i=1}^{n} i x_{i}^{2}\right)^{2}} \\
=\sum_{k=1}^... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,065 |
5. Given that $f(x)$ and $g(x)$ are both increasing linear functions defined on $\mathbf{R}$, and $f(x)$ is an integer if and only if $g(x)$ is an integer. Prove: For all $x \in \mathbf{R}$, $f(x) - g(x)$ is an integer.
（Liu Shixiong, problem contributor） | 5. Otherwise, by symmetry, we may assume $a>c$.
From $f\left(-\frac{b}{a}\right)=0$, we know that $g\left(-\frac{b}{a}\right)$ is an integer;
From $f\left(-\frac{b-1}{a}\right)=1$, we know that $g\left(-\frac{b-1}{a}\right)$ is an integer.
Thus, $g\left(-\frac{b}{a}\right)-g\left(-\frac{b-1}{a}\right)$
$$
\begin{array... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,066 |
6. As shown in Figure 2, in the acute triangle $\triangle ABC$, $AB > AC$, and $M$ is the midpoint of side $BC$. The external angle bisector of $\angle BAC$ intersects line $BC$ at point $P$. Points $K$ and $F$ are on line $PA$ such that $MF \perp BC$ and $MK \perp PA$. Prove that:
$BC^2 = 4 PF \cdot AK$.
(Supplied by ... | 6. As shown in Figure 4, let the circumcircle $\odot O$ of $\triangle ABC$ intersect line $FM$ at point $D$, and $AD$ intersects $BC$ at point $E$.
It is easy to know that $AD$ bisects $\angle BAC$.
Therefore, $AD \perp AP, AD \parallel MK$.
Thus, $\frac{MD}{FM} = \frac{AK}{FK}$.
Since $\angle FMC = \angle FAD = 90^{\c... | BC^2 = 4 PF \cdot AK | Geometry | proof | Yes | Yes | cn_contest | false | 723,067 |
7. Given a positive integer $n(n \geqslant 3)$. For any permutation $P=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ of $1,2, \cdots, n$, if $i<j<k$, then $x_{j}$ is said to be between $x_{i}$ and $x_{k}$ (for example, in the permutation $(1,3,2,4)$, 3 is between 1 and 4, and 4 is not between 1 and 2). Let the set $S=\left... | 7. First, prove by mathematical induction that: $m \leqslant 2^{n-1}$.
When $n=3$, by the condition, there must be a number (let's say 1) among 1, 2, 3 that does not lie between the other two numbers in any $P_{i}$.
Therefore, the permutations $(2,1,3)$ and $(3,1,2)$ cannot appear.
Thus, there are at most $3!-2=4$ p... | 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,068 |
8. Find the smallest odd number $a$ greater than 5 that satisfies the following conditions: there exist positive integers $m_{1}, n_{1}, m_{2}, n_{2}$, such that
$$
a=m_{1}^{2}+n_{1}^{2}, a^{2}=m_{2}^{2}+n_{2}^{2} \text {, }
$$
and $m_{1}-n_{1}=m_{2}-n_{2}$. | 8. From $261=15^{2}+6^{2}, 261^{2}=189^{2}+180^{2}$,
$15-6=189-180$,
we know that 261 has the property described in the problem.
Next, we prove that 261 is the smallest odd number greater than 5 with the property described in the problem, i.e., there is no such odd number between 5 and 261.
If not, then there exists a... | 261 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,069 |
Example 7 The set of all real solutions to the equation $[2 x]+[3 x]=9 x-\frac{7}{4}$ is $\qquad$ . | According to the properties of the Gaussian function, we have
$$
2 x-1<[2 x] \leqslant 2 x, 3 x-1<[3 x] \leqslant 3 x \text {. }
$$
Therefore, $5 x-2<[2 x]+[3 x] \leqslant 5 x$.
Thus, $5 x-2<9 x-\frac{7}{4} \leqslant 5 x$.
Solving this, we get $-\frac{1}{16}<x \leqslant \frac{7}{16}$.
Hence, $-\frac{37}{16}<9 x-\frac{... | x=-\frac{1}{36}, \frac{7}{36} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,070 |
One. (25 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=2, a_{n}=2^{2 n} a_{n-1}+2^{n^{2}} n \quad (n=2,3, \cdots) \text {. }
$$
Find the general term $a_{n} (n=1,2, \cdots)$.
(Wu Shuxun, problem contributor) | Let $b_{n}=\frac{a_{n}}{2^{n^{2}}}$. Then $b_{n}=2 b_{n-1}+n$, i.e.,
$$
b_{n}+n=2\left(b_{n-1}+n-1\right)+2, b_{1}=1 \text {. }
$$
Let $c_{n}=b_{n}+n$. Then $c_{n}=2 c_{n-1}+2$, i.e.,
$$
\begin{array}{l}
c_{n}+2=2\left(c_{n-1}+2\right), c_{1}=2 \\
\Rightarrow c_{n}+2=2^{n-1}\left(c_{1}+2\right)=2^{n+1} \\
\Rightarrow ... | a_{n}=2^{n^{2}}\left(2^{n+1}-n-2\right) | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,071 |
II. (25 points) Given that $P A$ and $P B$ are tangents to $\odot O$, with points of tangency at $A$ and $B$ respectively, and $P C D$ is a secant of $\odot O$. A line through point $C$ parallel to $P A$ intersects chord $A B$ and $A D$ at points $E$ and $F$ respectively. Prove that $C E = E F$.
(Li Xinhuan, problem co... | As shown in Figure 1, let $G$ be the midpoint of chord $CD$, and connect $OG$, $BG$, $EG$, and $BC$. Then $OG \perp CD$. Therefore, points $A$, $B$, and $G$ all lie on the circle with diameter $OP$.
Thus, $\angle APG = \angle ABG$.
Since $CF \parallel AP$, we have $\angle APG = \angle ECG$.
Therefore, $\angle ECG = \an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,072 |
Three. (25 points) Find all positive integer triples $(x, y, z)$ such that $1+2^{x} \times 3^{y}=5^{z}$ holds.
(Zhang Lei) | Three, taking the original equation modulo 3 yields that $z$ is even.
Let $z=2 r\left(r \in \mathbf{N}_{+}\right)$. Then
$\left(5^{r}-1\right)\left(5^{r}+1\right)=2^{x} \times 3^{y}$.
Since $\left(5^{\prime}-1,5^{\prime}+1\right)=\left(5^{\prime}-1,2\right)=2$, we have
$\left(\frac{5^{r}-1}{2}, \frac{5^{\prime}+1}{2}\r... | (3,1,2) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,073 |
Four. (25 points) In a $7 \times 7$ grid, there are 64 intersection points (referred to as "nodes") where chess pieces can be placed, with at most 1 piece per point, totaling $k$ pieces. If no matter how the pieces are placed, there always exist 4 pieces such that the nodes they occupy form the four vertices of a recta... | First, as shown in Figure 2, when $k=24$, it is possible that no 4 nodes with chess pieces form the four vertices of a rectangle, so $k \geqslant 25$.
Second, we will prove that when $k=25$, there must exist 4 nodes with chess pieces that form the four vertices of a rectangle.
If not, consider the row (or column) wit... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,074 |
Five. (25 points) Let positive real numbers $a$, $b$, $c$ satisfy
$$
(a+2 b)(b+2 c)=9 \text {. }
$$
Prove: $\sqrt{\frac{a^{2}+b^{2}}{2}}+2 \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \geqslant 3$.
(Zhang Lei) | Because $a, b, c > 0$, so,
$$
\begin{array}{l}
\sqrt{\frac{a^{2}+b^{2}}{2}}+2 \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \\
\geqslant \frac{a+b}{2}+2 \cdot \frac{b+c}{2} \\
=\frac{1}{2}[(a+2 b)+(b+2 c)] \\
\geqslant \sqrt{(a+2 b)(b+2 c)}=3 .
\end{array}
$$ | 3 | Inequalities | proof | Yes | Yes | cn_contest | false | 723,075 |
Six. (25 points) Given that $\odot O$ is the incircle of $\triangle A B C$, $D$, $E$, and $N$ are the points of tangency. Connect $N O$ and extend it to intersect $D E$ at point $K$. Connect $A K$ and extend it to intersect $B C$ at point $M$. Prove that $M$ is the midpoint of $B C$.
(Kang Chunbo, problem contributor) | Six, as shown in Figure 3, connect $O D$ and $O E$. Then $O, D, B, N$ are concyclic. Therefore,
$$
\begin{array}{l}
\angle K O D = \angle B, \angle K O E = \angle C. \\
\text { Hence } \frac{D K}{K E} = \frac{S_{\triangle O D K}}{S_{\triangle O E K}} = \frac{O D \cdot O K \sin \angle D O K}{O E \cdot O K \sin \angle K ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,076 |
Seven, (25 points) Find
$$
[x, y, z]=(x, y)+(y, z)+(z, x)
$$
satisfying $x \leqslant y \leqslant z,(x, y, z)=1$ for all positive integer solutions, where $[m, n]$ and $(m, n)$ represent the least common multiple and greatest common divisor of positive integers $m, n$, respectively. | Seven, because $(x, y, z)=1$, so,
$\frac{(x, y)}{[x, y, z]}=\frac{(x, y)}{[[x, y], z]}$
$\leqslant \frac{(x, y)}{[(x, y), z]}=\frac{(x, y)}{(x, y) z}=\frac{1}{z}$.
Similarly, $\frac{(x, z)}{[x, y, z]} \leqslant \frac{1}{x}, \frac{(z, x)}{[x, y, z]} \leqslant \frac{1}{y}$.
By the given condition, $\frac{1}{x}+\frac{1}{y... | (x, y, z)=(1,1,3),(1,2,4),(2,3,6) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,077 |
Eight, let $x, y, z \in [0,1]$, and
$$
|x-y| \leqslant \frac{1}{2}, |y-z| \leqslant \frac{1}{2}, |z-x| \leqslant \frac{1}{2} \text{. }
$$
Find the minimum and maximum values of $W=x+y+z-xy-yz-zx$.
(Liu Kangning, An Zhenping) | Because $x, y, z \in [0,1]$, we have $x+y+z \geqslant xy + yz + zx \Rightarrow W \geqslant 0$.
When $x=y=z=0$ or 1, $W_{\text{min}}=0$.
By the pigeonhole principle, among $x, y, z$, there must be two that are both no greater than $\frac{1}{2}$ or both no less than $\frac{1}{2}$.
Since the problem remains unchanged whe... | \frac{5}{6} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,078 |
1. Given that $a$ and $b$ are distinct positive real numbers, and $a^{3}-b^{3}$ $=a^{2}-b^{2}$. Then the range of values for $a+b$ is ( ).
(A) $0<a+b<\frac{1}{3}$
(B) $1<a+b<\frac{4}{3}$
(C) $\frac{3}{4}<a+b<1$
(D) $\frac{1}{3}<a+b<2$ | - 1. B.
From the given, it is easy to get $a^{2}+a b+b^{2}=a+b$, which means $\frac{a^{2}+a b+b^{2}}{a+b}=1$.
Since $(a+b)^{2}>a^{2}+a b+b^{2}$, therefore, $a+b>\frac{a^{2}+a b+b^{2}}{a+b}=1$.
Also, $(a-b)^{2}>0(a \neq b)$, then $3(a+b)^{2}<4\left(a^{2}+a b+b^{2}\right)$,
which means $a+b<\frac{4}{3}\left(\frac{a^{2}+... | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,079 |
2. Let $D$ be any point on the side $A B$ of $\triangle A B C$ (excluding points $A$ and $B$). $D$ moves parallel to $B C$ to point $E$ on side $A C$, then from $E$ moves parallel to $A B$ to point $F$ on side $B C$; then from $F$ moves parallel to $C A$ to point $G$ on side $A B$; and so on. A movement along a line pa... | 2. C.
According to the problem:
(1) When point $D$ starts from the midpoint of side $AB$, points $E$ and $F$ are the midpoints of $AC$ and $BC$, respectively. Therefore, such a point translation transformation requires only three times (minimum).
So, $n=3$.
(2) As shown in Figure 4, when point $D$ starts from any poin... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,080 |
Example 8 The number of solutions to the equation $x^{2}-2[x]-3=0$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | From the given equation, we get $[x]=\frac{x^{2}-3}{2}$.
And $x-1<[x] \leqslant x$, so $x-1<\frac{x^{2}-3}{2} \leqslant x$.
Solving this, we get $-1 \leqslant x<1-\sqrt{2}$ or $1+\sqrt{2}<x \leqslant 3$.
Therefore, $[x]$ can only take the values $-1,2,3$.
When $[x]=-1$, $\frac{x^{2}-3}{2}=-1$, solving this gives $x=-1$... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,081 |
3. Given $x-\frac{a y z}{x^{2}}=y-\frac{a z x}{y^{2}}=z-\frac{a x y}{z^{2}}=\lambda$, and the real numbers $x, y, z$ are distinct. Then $x+y+z=(\quad)$.
(A) $\frac{\lambda}{a}$
(B) $a \lambda$
(C) $\lambda-a$
(D) $\lambda+a$ | 3. D.
From the given, we have
$$
x^{3}-a y z=\lambda x^{2}, \quad y^{3}-a z x=\lambda y^{2} \text {. }
$$
Subtracting the two equations, we get
$$
x^{3}-y^{3}+a z(x-y)=\lambda\left(x^{2}-y^{2}\right) \text {. }
$$
Since $x-y \neq 0$, we have
$$
\lambda(x+y)=x^{2}+x y+y^{2}+a z \text {. }
$$
Similarly, $\lambda(y+z)... | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,082 |
4. Let the three medians $A D, B E, C F$ of $\triangle A B C$ intersect at point $G$. If $B C=a, C A=b, A B=c$, then the value of $A G^{2}+$ $B G^{2}+C G^{2}$ is ( ).
(A) $\frac{1}{3}\left(a^{2}+b^{2}+c^{2}\right)$
(B) $\frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)$
(c) $\frac{1}{3}(a b+b c+c i)$
(D) $\frac{2}{3}(a b+b c+c... | 4. A.
From the median length formula of a triangle, we have
$$
\begin{array}{l}
2\left(A G^{2}+B G^{2}\right)=A B^{2}+(2 F G)^{2} \\
=A B^{2}+C G^{2} .
\end{array}
$$
Similarly, $2\left(B G^{2}+C G^{2}\right)=B C^{2}+A G^{2}$,
$$
2\left(C G^{2}+A G^{2}\right)=A C^{2}+B G^{2} \text {. }
$$
Adding the three equations ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,083 |
5. Given real numbers $a, b, c, d$ satisfy
$$
a^{2}+b^{2}=1, c^{2}+d^{2}=1, a c+b d=0 \text {. }
$$
The following conclusions are given:
(1) $a^{2}+c^{2}=1$;
(2) $b^{2}+d^{2}=1$;
(3) $a b+c d=0$;
(4) $a d+b c=0$.
The number of correct conclusions is ( ) .
(A) 1
(B) 2
(C) 3
(D) 4 | 5. C.
Notice
$$
\begin{array}{l}
a^{2}+b^{2}=c^{2}+d^{2}=1 \\
\Rightarrow\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \\
\quad=a^{2} c^{2}+a^{2} d^{2}+b^{2} c^{2}+b^{2} d^{2}=1 ; \\
a c+b d=0 \Rightarrow a^{2} c^{2}=b^{2} d^{2} .
\end{array}
$$
Then $a^{2} d^{2}+2 a^{2} c^{2}+b^{2} c^{2}=1$, that is
$$
a^{2}\left... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,084 |
6. Given that the sides of $\square A B C D$ are $A B=m, A D=n$ $(m>n)$. If $P$ is a moving point on side $C D$, and let $D P=x$, the line $A P$ intersects the extension of $B C$ at point $Q$, such that $D P+C Q$ is minimized, then the relationship that $m$, $n$, and $x$ should satisfy is ( ).
(A) $x=\frac{m+n}{2}$
(B)... | 6. B.
As shown in Figure 5, we have
$$
\begin{array}{l}
\triangle C P Q \backsim \triangle D P A \\
\Rightarrow \frac{C Q}{D A}=\frac{C P}{D P} \\
\Rightarrow \frac{C Q}{n}=\frac{m-x}{x} \\
\Rightarrow C Q=\frac{m-x}{x} \cdot n .
\end{array}
$$
Let's assume $y=D P+C Q$. Then
$$
y=x+\frac{n(m-x)}{x}=x+\frac{m n}{x}-n ... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,085 |
1. $\left(\frac{1+\sqrt{5}}{2}\right)^{6}+\left(\frac{1-\sqrt{5}}{2}\right)^{6}=$ | Let $x_{1}=\frac{1+\sqrt{5}}{2}, x_{2}=\frac{1-\sqrt{5}}{2}$. Then $x_{1}+x_{2}=1, x_{1} x_{2}=-1$.
Thus, $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=3$.
Therefore, $x_{1}^{6}+x_{2}^{6}=\left(x_{1}^{2}\right)^{3}+\left(x_{2}^{2}\right)^{3}$
$=\left(x_{1}^{2}+x_{2}^{2}\right)\left[\left(x_{1}^{2}\rig... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,086 |
3. For any real value of $t$, the quadratic function
$$
y=\left(t^{2}+t+1\right) x^{2}-2(a+t)^{2} x+\left(t^{2}+3 a t+b\right)
$$
passes through the point $(1,0)$. Then $a=$ $\qquad$ ,$b=$ $\qquad$ | 3. 1,1 .
Let $t=0,1$, then we have
$$
\begin{array}{l}
y=x^{2}-2 a^{2} x+b, \\
y=3 x^{2}-2(a+1)^{2} x+(1+3 a+b) .
\end{array}
$$
According to the problem, the graphs of functions (1) and (2) both pass through the point $(1,0)$, so,
$$
\begin{array}{l}
1-2 a^{2}+b=0, \\
3-2(a+1)^{2}+(1+3 a+b)=0 .
\end{array}
$$
Solv... | a=1, b=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,088 |
One, (20 points) Given that $a$ and $b$ are real numbers, the equation $\left|x^{2}+a x+b\right|=2$ has three distinct real roots.
(1) Prove that $a^{2}-4 b=8$;
(2) If the three distinct real roots of the equation are exactly the degree measures of the three interior angles of a triangle, prove that the triangle must h... | (1) From the original equation, we get
$$
x^{2}+a x+b-2=0
$$
or $x^{2}+a x+b+2=0$,
with discriminants respectively
$$
\Delta_{1}=a^{2}-4 b+8, \Delta_{2}=a^{2}-4 b-8 .
$$
Since the original equation has three distinct real roots, and it is evident that equations (1) and (2) have no common roots, one of the equations m... | a=-16, b=62 | Algebra | proof | Yes | Yes | cn_contest | false | 723,090 |
II. (25 points) As shown in Figure 3, given that $\odot O_{1}$ and $\odot O_{2}$ are disjoint, a tangent line from point $O_{1}$ to $\odot O_{2}$ (with $T_{2}$ as the point of tangency) intersects $\odot O_{1}$ at points $A_{1}$ and $A_{2}$; a tangent line from point $O_{2}$ to $\odot O_{1}$ (with $T_{1}$ as the point ... | II. As shown in Figure 7, connect $O_{1} T_{1}$, $O_{2} T_{2}$, $T_{1} T_{2}$, $A_{1} T_{1}$, $A_{2} T_{1}$, $B_{1} T_{2}$, $B_{2} T_{2}$, $O_{1} O_{2}$.
It is easy to see that $\angle O_{1} T_{1} O_{2} = \angle O_{2} T_{2} O_{1} = 90^{\circ}$.
Therefore, $T_{1}$, $T_{2}$, $O_{2}$, and $O_{1}$ are concyclic.
Thus, $\an... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,091 |
Example 9 Let $a$ be a rational number. Among the following four conclusions, the correct one is ( ).
(A) $[a]+[-a]=0$
(B) $[a]+[-a]$ equals 0 or 1
(C) $[a]+[-a] \neq 0$
(D) $[a]+[-a]$ equals 0 or -1
(17th "Hope Cup" Mathematics Invitational (Grade 7)) | (1) If $a$ is an integer, then $[a]=a$, and
$[-a]=-a$.
Therefore, $[a]+[-a]=a+(-a)=0$.
(2) If $a$ is not an integer, then $a-1<[a]<a$, and $-a-1<[-a]<-a$.
Therefore, $-2<[a]+[-a]<0$.
Thus, $[a]+[-a]=-1$.
Combining (1) and (2), the correct choice is (D).
[Note] The above solution uses the method of case analysis to prov... | D | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,092 |
Three, (25 points) If the sum, difference, product, and quotient of two unequal natural numbers add up to a perfect square, then such two numbers are called a "wise pair" (for example, $(8,2)$ is a wise pair, since $\left.(8+2)+(8-2)+8 \times 2+\frac{8}{2}=36=6^{2}\right)$.
If both of these natural numbers do not exce... | Three, let $(a, b)$ be a wise array, and without loss of generality, assume $a > b$.
By definition, we have
$$
(a+b)+(a-b)+a b+\frac{a}{b}=m^{2}
$$
$(a, b, m$ are natural numbers), which simplifies to
$$
2 a+a b+\frac{a}{b}=m^{2}.
$$
Since $2 a, a b, m^{2}$ are all natural numbers, $\frac{a}{b}$ must also be a natural... | 53 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,093 |
1. Given the function $f(x)=\log _{a}\left(a x^{2}-x+\frac{1}{2}\right)$ is always positive on the interval $[1,2]$. Then the range of the real number $a$ is | 1. $\frac{1}{2}\frac{3}{2}$.
From $a x^{2}-x+\frac{1}{2}>0$, we get
$$
a>\frac{x-\frac{1}{2}}{x^{2}}=h(x) \text {. }
$$
Since $h^{\prime}(x)=\frac{1-x}{x^{3}}$, when $1 \leqslant x \leqslant 2$, $h^{\prime}(x)\frac{1}{2}$, and $a \neq 1$.
When $\frac{1}{2}0$, we know $\frac{1}{2}1$, $f(x)$ is an increasing function on... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,094 |
2. Let positive real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}=1$. Then the maximum value of $s=a^{2} b c+a b^{2} c+a b c^{2}$ is $\qquad$ | 2. $\frac{1}{3}$.
By the AM-GM inequality and Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
a^{2} b c+a b^{2} c+a b c^{2}=a b c(a+b+c) \\
\leqslant \sqrt{\left(\frac{a^{2}+b^{2}+c^{2}}{3}\right)^{3}} \cdot \sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}=\frac{1}{3} .
\end{array}
$$
Equality holds if and only if $a=b=... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,095 |
3. Let $\alpha$, $\beta$, and $\gamma$ be the angles formed by the diagonal of a rectangular prism with the three faces sharing a common vertex. Then the range of $\alpha+\beta+\gamma$ is $\qquad$ . | 3. $\left(\frac{\pi}{2}, 3 \arcsin \frac{\sqrt{3}}{3}\right]$.
In a rectangular parallelepiped, there is
$$
\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=1 \text {. }
$$
Notice that
$$
\begin{array}{l}
\sin ^{2} \alpha=1-\sin ^{2} \beta-\sin ^{2} \gamma=\cos ^{2} \beta-\sin ^{2} \gamma \\
=\frac{1}{2}(1+\cos 2 \b... | \left(\frac{\pi}{2}, 3 \arcsin \frac{\sqrt{3}}{3}\right] | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,096 |
5. Let the function $f(n)$ be defined on the set of positive integers, for any positive integer $n$, we have $f(f(n))=4 n+3$, and for any non-negative integer $k$, we have
$$
f\left(2^{k}\right)=2^{k+1}+1 \text {. }
$$
Then $f(2303)=$ | 5.4607.
Notice
$$
\begin{array}{l}
2303=3+4 \times 3+4^{2} \times 3+4^{3} \times 3+4^{4} \times 2^{3} \text {. } \\
\text { And } f(4 n+3)=f(f(f(n)))=4 f(n)+3 \text {, then } \\
f(2303) \\
=3+4 f\left(3+4 \times 3+4^{2} \times 3+4^{3} \times 2^{3}\right) \\
=\cdots \\
=3+4 \times 3+4^{2} \times 3+4^{3} \times 3+4^{4} ... | 4607 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,098 |
6. As shown in Figure 1, let $G$ and $H$ be the centroid and orthocenter of $\triangle ABC$ respectively, $F$ be the midpoint of segment $GH$, and the circumradius of $\triangle ABC$ be $R=1$. Then
$$
\begin{array}{l}
|\overrightarrow{A F}|^{2}+|\overrightarrow{B F}|^{2}+|\overrightarrow{C F}|^{2} \\
= \\
.
\end{array}... | 6.3.
Let the circumcenter $O$ of $\triangle ABC$ be the origin of a Cartesian coordinate system. Then,
$$
\begin{array}{l}
\overrightarrow{O H}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}, \\
\overrightarrow{O G}=\frac{1}{3}(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C})
\end{array}... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,099 |
7. Let point $P$ move on the right branch of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ excluding the vertices, $E$ and $F$ be its left and right foci, respectively. Let $A$ be the excenter of $\triangle P E F$ in $\angle P E F$. Then the equation of the trajectory of point $A$ is $\qquad$ | 7. $\frac{x^{2}}{25}-\frac{9 y^{2}}{25}=1(x>5)$.
From the problem, we know that the semi-major axis length $a=4$, and the semi-focal distance $c=5$.
Let $A(x, y), P\left(x_{0}, y_{0}\right)$. Then $x_{0}>4$.
Let $P A$ intersect the $x$-axis at point $B\left(x_{B}, y_{B}\right)$.
By the angle bisector property, we have... | \frac{x^{2}}{25}-\frac{9 y^{2}}{25}=1(x>5) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,100 |
8. The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=1$, and for $n \geqslant 2$, $a_{n}=\left\{\begin{array}{ll}a_{\frac{n}{2}}+1, & n \text { is even; } \\ \frac{1}{a_{n-1}}, & n \text { is odd. }\end{array}\right.$ If $a_{n}=\frac{20}{11}$, then the positive integer $n=$ $\qquad$ | 8. 198 .
From the problem, we know that when $n$ is even, $a_{n}>1$; when $n(n>1)$ is odd, $a_{n}=\frac{1}{a_{n-1}}1$, so, $n$ is even. Thus, $a_{\frac{n}{2}}=\frac{20}{11}-1=\frac{9}{11}1, \frac{n}{2}-1$ is even;
$a_{\frac{n-2}{4}}=\frac{11}{9}-1=\frac{2}{9}1, \frac{n-6}{4}$ is even;
$a_{\frac{n-6}{8}}=\frac{9}{2}-1=... | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,101 |
$\begin{array}{l}\text { 9. (16 points) Let } \\ f_{n}(x)=\sin ^{2 n} x+\cos ^{2 n} x(x \in \mathbf{R}, n \in \mathbf{N}) \text {. } \\ \text { Let } S=f_{1}(x)-f_{2}(x)+f_{3}(x)-f_{4}(x)+f_{5}(x) \text {. }\end{array}$
Find the maximum and minimum values of $S$. | $$
\begin{array}{l}
f_{n}(x)=\sin ^{2 n} x+\cos ^{2 n} x \\
=f_{n-1}(x)-\sin ^{2} x \cdot \cos ^{2} x \cdot f_{n-2}(x) .
\end{array}
$$
Let $\sin ^{2} x \cdot \cos ^{2} x=k$. Then
$$
f_{n}(x)=f_{n-1}(x)-k f_{n-2}(x) \text {, }
$$
or $f_{n-1}(x)-f_{n}(x)=k f_{n-2}(x)$.
It is easy to see that $f_{0}(x)=2, f_{1}(x)=1$,
... | S_{\max }=1, S_{\min }=\frac{11}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,102 |
10. (20 points) In $\triangle A B C$, if $\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\sqrt{3}$,
prove that $\triangle A B C$ has at least one angle of $60^{\circ}$. | 10. From $\sqrt{3}=\tan 60^{\circ}=\frac{\sin 60^{\circ}}{\cos 60^{\circ}}$, we know
$$
\begin{array}{l}
\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=\frac{\sin 60^{\circ}}{\cos 60^{\circ}} . \\
\text { Hence } \sin \left(A-60^{\circ}\right)+\sin \left(B-60^{\circ}\right)+\sin \left(C-60^{\circ}\right) \\
=0 .
\en... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,104 |
11. (20 points) If there are two non-coincident circles $\Gamma_{1}$ and $\Gamma_{2}$ in a plane, does there exist a point $P$ such that there are infinitely many pairs of perpendicular lines $l_{1}$ and $l_{2}$ passing through $P$, and these lines intersect circles $\Gamma_{1}$ and $\Gamma_{2}$ respectively, with the ... | 11. As shown in Figure 3, establish a Cartesian coordinate system such that the centers of circles $\Gamma_{1}$ and $\Gamma_{2}$ are $O_{1}(-a, 0)$ and $O_{2}(a, 0)$, respectively. Let $r_{1}$ and $r_{2}$ be the radii of circles $\Gamma_{1}$ and $\Gamma_{2}$, and let point $P(m, n)$. Then the equations of lines $l_{1}$... | not found | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,105 |
One, (40 points) As shown in Figure 2, the extensions of the two pairs of opposite sides of the convex quadrilateral $A B D F$ intersect at points $C$ and $E$ respectively. Let $O_{1}, O_{2}, O_{3}, O_{4}$ be the circumcenters of $\triangle A C F$, $\triangle B C D$, $\triangle D E F$, and $\triangle A B E$ respectivel... | As shown in Figure 4.
Let the circumcircle $\odot O_{2}$ of $\triangle B C D$ and the circumcircle $\odot O_{3}$ of $\triangle D E F$ intersect at another point $M$ (otherwise, if it coincides with point $D$, it implies that $B C$ is parallel to $F E$, which is a contradiction). The projections of $M$ on lines $A C$, $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,106 |
Sure, here is the translated text:
```
II. (40 points) Let $a, b, c$ be non-negative real numbers. Prove:
$$
\sum \frac{1}{a^{2}+2 b c+3 a(b+c)} \leqslant \frac{1}{a b+b c+c a},
$$
where, “ ${ }^{4}$ ” denotes the cyclic sum.
``` | Let $a+b+c=A, ab+bc+ca=B$. Then the original inequality is equivalent to
$$
\begin{array}{l}
\sum \frac{B}{aA+2B} \leqslant 1 \\
\Leftrightarrow \sum\left(\frac{-B}{aA+2B}+\frac{1}{2}\right) \geqslant -1+\frac{3}{2} \\
\Leftrightarrow \sum \frac{aA}{aA+2B} \geqslant 1. \\
\end{array}
$$
By the Cauchy-Schwarz inequalit... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,107 |
Four. (50 points) In a plane, there are 12 points, no three of which are collinear. Each pair of points is connected by a line segment (or edge). These line segments are colored in red and blue, with each line segment being colored exactly one color. Among them, the number of red line segments starting from a certain p... | Four, Solution 1 Notice that the number of red line segments drawn from each point can only be one of the values $0,1, \cdots, 11$, and $0$ and $11$ cannot appear simultaneously. Therefore, there are two possible scenarios:
(1) $0,1, \cdots, 10$;
(2) $1,2, \cdots, 11$.
If it is scenario (2), when point $A$ draws $2 k+1... | 40, 55 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,109 |
Given a monic quartic equation
$$
x^{4}+b x^{3}+c x^{2}+d x+e=0 \text {, }
$$
where $b, c, d, e$ are all integers, and $(b+c+d) e$ is an odd number. Prove: This monic quartic equation has no integer roots. | Proof by contradiction.
Assume the equation has an integer root $x_{0}$. Then
$$
\begin{array}{l}
f(x)=x^{4}+b x^{3}+c x^{2}+d x+e \\
=\left(x-x_{0}\right)\left(x^{3}+p x^{2}+q x+r\right),
\end{array}
$$
where $p, q, r$ are integers,
$$
x_{0} r=-e .
$$
Since $(b+c+d) e$ is odd, $b+c+d$ is odd, and $e$ is also odd.
Fr... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,110 |
Initial 284 As shown in Figure $2, \odot O_{1}$ and $\odot O_{2}$ intersect at points $M$ and $N, M A$ is a tangent to $\odot O_{2}$ intersecting $\odot O_{1}$ at point $A, M B$ is a tangent to $\odot O_{1}$ intersecting $\odot O_{2}$ at point $B$, line $A B$ intersects $\odot O_{1}$ and $\odot O_{2}$ at points $C$ and... | Proof As shown in Figure 2, let the extension of $MN$ intersect the circumcircle of $\triangle MAB$ at point $E$, and connect $AE, AN, BE, BN$.
From $\angle MAE + \angle MBE = 180^{\circ}$, we get
$\sin \angle MAE = \sin \angle MBE$.
Obviously, $\frac{AP}{BP} = \frac{S_{\triangle MME}}{S_{\triangle MBE}} = \frac{AM \cd... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,111 |
283 Find the unit digit of $\left(\frac{5+\sqrt{21}}{2}\right)^{2010}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solve the problem, which is to find the remainder when $\left[\left(\frac{5+\sqrt{21}}{2}\right)^{2010}\right]$ is divided by 10.
Let $a_{n}=\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\left(\frac{5-\sqrt{21}}{2}\right)^{n}\left(n \in \mathbf{N}_{+}\right)$.
Since $\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5$,
$$
\frac{5+... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,112 |
Given seven points on a circle, connect each pair of points. Find the minimum number of intersection points of these lines inside the circle. | We prove that these lines intersect at most three sets of three lines at a point inside the circle, and it is obvious that there cannot be four lines intersecting at a point.
First, select seven vertices of a regular octagon,
it is clear that there are three sets of three lines intersecting at a point (as shown in Figu... | 29 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,113 |
Example 1 In a regular 2004-gon $A_{1} A_{2} \cdots A_{2004}$, fill each vertex with one of the numbers $1,2, \cdots, 501$ arbitrarily. Prove that there must exist four vertices satisfying the following conditions:
(1) The quadrilateral formed by these vertices is a rectangle;
(2) The sum of the numbers filled at the o... | Proof (1) From the problem, we know that vertices $A_{i}$ and $A_{i+1002}$ $(i=1,2, \cdots, 1002)$ form a pair of points symmetric about the center. Thus, the 2004 vertices can be divided into 1002 pairs.
Connecting the vertices of each pair in sequence, we can obtain a quadrilateral. Since its diagonals bisect each o... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,114 |
Example 2: 10 million points are randomly given on a plane. Can a straight line be used to separate them so that there are 5 million points on each side of the line?
$(1985$, China University of Science and Technology Young Class Admission Examination for Reinterview) | Consider the 10 million points. Among the line segments connecting any two of these points, there must be two points with the maximum distance (let's call them $M$ and $N$). Draw a circle with $M$ as the center and a radius greater than the distance $MN$. Then, all 10 million points are inside this circle.
Since the n... | proof | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,115 |
5. A square with a side length of 5 covers 10 circular paper pieces with a radius of 1. Prove: There are always two circular paper pieces, the overlapping area of which is not less than $\frac{\pi}{2}-1$. | As shown in Figure 9, the centers of 10 circles can only fall within the 9 central squares (the thick-lined part). Therefore, there must be two circle centers that fall within the same small square, and the maximum distance between the two centers is $\sqrt{2}$. Hence, the overlapping area of the two circles is
$$
\beg... | \frac{\pi}{2}-1 | Geometry | proof | Yes | Yes | cn_contest | false | 723,118 |
Example 1 Given $\triangle A B C$. If for any $t \in \mathbf{R}$, $|\overrightarrow{B A}-t \overrightarrow{B C}| \geqslant|\overrightarrow{A C}|$, then $\triangle A B C$ must be ( ).
(A) acute triangle
(B) obtuse triangle
(C) right triangle
(D) uncertain
(2006, National High School Mathematics Competition) | 【Analysis】Convert the vector relationship of the condition into geometric characteristics.
Solution Let $t \overrightarrow{B C}=\overrightarrow{B D}$.
Then point $D$ lies on the line
$B C$, as shown in Figure 1. Therefore,
$$
\begin{aligned}
\overrightarrow{B A}-t \overrightarrow{B C} \\
=\overrightarrow{B A}-\overrigh... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,119 |
Example 2 Suppose point $O$ is inside $\triangle A B C$, and $\overrightarrow{O A}+2 \overrightarrow{O B}+3 \overrightarrow{O C}=0$.
Then $S_{\triangle A B C}: S_{\triangle A O C}=(\quad$.
(A) 2
(B) $\frac{3}{2}$
(C) 3
(D) $\frac{5}{3}$
(2004, National High School Mathematics Competition) | Let $e$ be the unit normal vector of $A C$, and construct $B D \perp A C$ at point $D$, and $O E \perp A C$ at point $E$, as shown in the figure.
2. Then
$$
\begin{aligned}
D B & =\overrightarrow{A B} \cdot e \\
& =(\overrightarrow{A O}+\overrightarrow{O B}) \cdot e \\
& =\left(\overrightarrow{A O}+\frac{\overrightarro... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,120 |
Example 3 Given a convex quadrilateral $A B C D, A D$ intersects $B C$ at point $E, A C$ intersects $B D$ at point $I$. Prove: The centroid of $\triangle E D C$ coincides with the centroid of $\triangle I A B$ if and only if $A B \parallel C D$ and $I C^{2}=I A \cdot A C$. ${ }^{[1]}$
(2005, Romanian Mathematical Olymp... | $$
\begin{array}{l}
\overrightarrow{C I}=m \overrightarrow{I A}, \\
\overrightarrow{D I}=n \overrightarrow{I B},
\end{array}
$$
where, \( m, n \in \mathbf{R}_{+} \),
$$
\begin{array}{l}
\overrightarrow{I E} \\
=a \overrightarrow{I C}+(1-a) \overrightarrow{I B} \\
=b \overrightarrow{I D}+(1-b) \overrightarrow{I A} \tex... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,121 |
Example 4 Let $K, M$ be two points on the side $AB$ of $\triangle ABC$, and $L, N$ be two points on the side $AC$, with $K$ between $M, B$ and $L$ between $N, C$, and $\frac{BK}{KM}=\frac{CL}{LN}$. Prove that the orthocenters of $\triangle ABC$, $\triangle AKL$, and $\triangle AMN$ are collinear.
(2006, China National ... | As shown in Figure 4, let the orthocenters of $\triangle A B C$, $\triangle A K L$, and $\triangle A M N$ be $H_{1}$, $H_{2}$, and $H_{3}$, respectively. Draw a perpendicular from $M$ to $A C$, intersecting the line $H_{1} H_{2}$ at point $H_{3}^{\prime}$. Draw a perpendicular from $N$ to $A B$, intersecting the line $... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,122 |
Example 5 As shown in Figure 5, given that $H$ is the orthocenter of acute $\triangle A B C$, a circle with the midpoint of side $B C$ as its center and passing through point $H$ intersects line $B C$ at points $A_{1}$ and $A_{2}$. A circle with the midpoint of side $C A$ as its center and passing through point $H$ int... | 【Analysis】If these six points are concyclic, then the center of the circle is the intersection of the perpendicular bisectors of $A_{1} A_{2}$, $B_{1} B_{2}$, and $C_{1} C_{2}$, which is the circumcenter $O$ of $\triangle A B C$. Noting the symmetry of these six points, it is sufficient to prove that $O A_{1}$ is a con... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,123 |
Example 6 Given that the diagonals of a convex quadrilateral $ABCD$ intersect at point $M, P, Q$ are the centroids of $\triangle AMD, \triangle CMB$ respectively, $R, S$ are the orthocenters of $\triangle DMC, \triangle MAB$ respectively. Prove that $PQ \perp RS$.
(2003, China National Training Team Problem) | $$
\begin{array}{l}
\overrightarrow{O P}=\frac{1}{3}(\overrightarrow{O A}+\overrightarrow{O M}+\overrightarrow{O D}), \\
\overrightarrow{O Q}=\frac{1}{3}(\overrightarrow{O C}+\overrightarrow{O M}+\overrightarrow{O B}), \\
\overrightarrow{M R} \cdot \overrightarrow{D C}=\overrightarrow{C R} \cdot \overrightarrow{M B}=\o... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,124 |
Example 7 Let circle $\Gamma$ be the circumcircle of $\triangle ABC$, and let $P$ be a point inside $\triangle ABC$. The rays $AP$, $BP$, and $CP$ intersect circle $\Gamma$ at points $A_1$, $B_1$, and $C_1$, respectively. Let the points symmetric to $A_1$, $B_1$, and $C_1$ with respect to the midpoints of sides $BC$, $... | 【Analysis】To prove that the circumcircle of $\triangle A_{2} B_{2} C_{2}$ passes through the orthocenter $H$ of $\triangle A B C$, we need to find the circumcenter $X_{0}$ of $\triangle A_{2} B_{2} C_{2}$ and prove that $X_{0} A_{2}=X_{0} H$.
Proof As shown in Figure 7, let the midpoints of $B C$, $C A$, and $A B$ be ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,125 |
Example 3 As shown in Figure $1, a / / b$, line $a$ has ten points $A_{1}, A_{2}$, $\cdots, A_{10}$, and line
$b$ has nine points $B_{1}, B_{2}, \cdots, B_{9}$
nine points. Connecting each point on
$a$ with each point on
$b$ can result in many line segments, given that no three line segments intersect at one point. How... | Solution: Take two points each on $a$ and $b$, the four points determine a unique intersection point. Taking two points from $a$ has $10 \times 9 \div 2=45$ methods, taking two points from $b$ has $9 \times 8 \div 2=36$ methods. In total, there are $45 \times 36=1620$ methods. | 1620 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,126 |
Example 8 Given a convex hexagon $A B C D E F$, any two opposite sides (such as $A B$ and $D E$, $B C$ and $E F$, $C D$ and $F A$, which are three pairs of opposite sides of the convex hexagon $A B C D E F$) have the following property: the distance between the midpoints of these sides is equal to $\frac{\sqrt{3}}{2}$ ... | Prove as follows:
8, Let $a=\overrightarrow{A B}, b=\overrightarrow{B C}, c=\overrightarrow{C D}, d=\overrightarrow{D E}, e=\overrightarrow{E F}, f=\overrightarrow{F A}$. And let $M, N$ be the midpoints of $A B, D E$, respectively. Then
$$
\begin{array}{l}
\overrightarrow{M N}=\frac{1}{2} a+b+c+\frac{1}{2} d, \\
\overr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,127 |
3. In $\triangle A B C$, $\angle A=60^{\circ}, A B>A C, O$ is the circumcenter, the two altitudes $B E$ and $C F$ intersect at point $H$, points $M$ and $N$ lie on segments $B H$ and $H F$ respectively, and satisfy $B M=C N$. Find the value of $\frac{M H+N H}{O H}$.
(2002, National High School Mathematics Competition) | From the known conditions, we have
$$
\begin{array}{l}
M H+N H=B H-C H \\
=2 R(\cos B-\cos C)=2 \sqrt{3} \sin \frac{C-B}{2},
\end{array}
$$
and \( |\overrightarrow{A H}|=R \) (where \( R \) is the radius of the circumcircle of \( \triangle A B C \)).
From \( \overrightarrow{O H}=\overrightarrow{O A}+\overrightarrow{A ... | \sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,128 |
4. Let $\triangle ABC$ be a non-right triangle, with orthocenter $H$, and $M_{1}$, $M_{2}$, $M_{3}$ be the midpoints of sides $BC$, $CA$, $AB$ respectively. Let $A_{1}$, $B_{1}$, $C_{1}$ be the points symmetric to $H$ with respect to $M_{1}$, $M_{2}$, $M_{3}$ respectively, and $A_{2}$, $B_{2}$, $C_{2}$ be the orthocent... | It is known that the orthocenter $A_{2}$ of $\triangle A_{1} B C$ and the orthocenter $A$ of $\triangle H B C$ are symmetric with respect to point $M_{1}$. Therefore,
$$
\begin{array}{l}
\overrightarrow{H B}+\overrightarrow{H C}=2 \overrightarrow{H M_{1}}=\overrightarrow{H A}+\overrightarrow{H A_{2}} \\
\Rightarrow \ov... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,129 |
Second question Let $k$ be a given positive integer,
$$
\begin{array}{l}
r=k+\frac{1}{2} . \\
\text { Let } f^{(1)}(r)=f(r)=r\lceil r\rceil, \\
f^{(l)}(r)=f\left(f^{(l-1)}(r)\right)(l \geqslant 2) .
\end{array}
$$
Prove: There exists a positive integer $m$, such that $f^{(m)}(r)$ is an integer, where $\lceil x\rceil$ ... | Proof 1 First, we use mathematical induction to prove:
$2 f^{(m)}(r) \in \mathbf{Z}$.
Obviously, $2 f^{(1)}(r)=(2 k+1)\lceil r\rceil \in \mathbf{Z}$.
When $2 f^{(m-1)}(r) \in \mathbf{Z}$,
$2 f^{(m)}(r)=2 f\left(f^{(m-1)}(r)\right)$
$=2 f^{(m-1)}(r)\left\lceil f^{(m-1)}(r)\right\rceil \in \mathbf{Z}$.
Therefore, equatio... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,131 |
Given an integer $n(n>2)$, let positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{k} \leqslant 1(k=1,2, \cdots, n)$, and let
$$
A_{k}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}(k=1,2, \cdots, n) .
$$
Prove: $\left|\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n} A_{k}\right|<\frac{n-1}{2}$. | Proof 1 Note that
$$
\begin{array}{l}
\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n} A_{k}=\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n} \frac{1}{k} \sum_{i=1}^{k} a_{i} \\
=\sum_{i=1}^{n} a_{i}-\sum_{i=1}^{n} a_{i} \sum_{k=1}^{n} \frac{1}{k} \\
=\sum_{i=1}^{n} a_{i}\left(1-\sum_{k=i}^{n} \frac{1}{k}\right) . \\
\text { And } \sum_{k=1}^{n... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,132 |
1. Given that $a$, $b$, and $c$ are all positive integers, and $a b c=$ 2010. Then the minimum value of $a+b+c$ is ( ).
(A) 84
(B) 82
(C) 78
(D) 76 | $$
\begin{array}{l}
\text { I.1.C. } \\
\text { From } 2010=2 \times 3 \times 5 \times 67=6 \times 5 \times 67 \\
=1 \times 30 \times 67=\cdots,
\end{array}
$$
we know that the minimum value of $a+b+c$ is 78. | 78 | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,134 |
2. When $a<1$, the simplified result of $\sqrt{-a^{3}(1-a)}$ is ( ).
(A) $a \sqrt{a(a-1)}$
(B) $-a \sqrt{a(a-1)}$
(C) $a \sqrt{a(1-a)}$
(D) $-a \sqrt{a(1-a)}$ | 2. B.
From the given, we know that $1-a>0$.
Also, since $-a^{3}(1-a) \geqslant 0$, it follows that $a \leqslant 0$.
Therefore, the original expression $=\sqrt{-a \cdot a^{2}(1-a)}$
$$
=|a| \cdot \sqrt{a(a-1)}=-a \sqrt{a(a-1)} .
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,135 |
3. When $x=\frac{1}{\sqrt{3}-2}$, the value of the function $f(x)=x^{3}+4 x^{2}-$ $2 x-6$ is ( ).
(A) $-3 \sqrt{3}$
(B) $3 \sqrt{3}$
(C) $-6 \sqrt{3}$
(D) $6 \sqrt{3}$ | 3. B.
Notice that $x=\frac{1}{\sqrt{3}-2}=-(2+\sqrt{3})$, i.e., $x+2=-\sqrt{3}$.
Therefore, when $x=\frac{1}{\sqrt{3}-2}$,
$$
\begin{array}{l}
f(x)=x(x+2)^{2}-6(x+1) \\
\quad=3 x-6(x+1)=-3(x+2)=3 \sqrt{3} .
\end{array}
$$ | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,136 |
Example: How many types of isosceles triangles with integer side lengths and a perimeter of 100 are there?
(8th "Hua Luogeng Cup" Junior Mathematics Invitational Final) | Let $a$, $b$, and $c$ be the lengths of the three sides of a triangle.
(1) $a=ba .\end{array}\right.$,
Solving, we get $25<a<33 \frac{1}{3}$.
Therefore, there are 8 possibilities.
(2) $a<b=c$.
In this case, $a=100-2 c$.
Thus, $\left\{\begin{array}{l}100-2 c<c \\ 1+2 c \leqslant 100\end{array}\right.$.
Solving, we get ... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,137 |
4. Find a parallelogram such that its two adjacent sides are $a$ and $2a$, and the acute angle between the two diagonals is $60^{\circ}$. Such a parallelogram should be ( ).
(A) only one
(B) two
(C) infinitely many
(D) none at all | 4. D.
As shown in Figure 4, intuitively, the locus of point $O$ for all $\triangle OBC$ with $BC=a$ as the base and the vertex angle $O$ as $60^{\circ}$ is the major arc $\overparen{BC}$ (as shown in Figure 4 (b)), excluding points $B$ and $C$. Among these, the maximum value of the median on the base $BC$ is the heigh... | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,138 |
5. Xiao Ming encountered a melon vendor on the street, hearing the vendor shout: “Large melons for 10 yuan each, medium melons for 10 yuan two, small melons for 10 yuan three. Pick and choose as you like.” Ming carefully examined the size and shape of the melons: the melons are nearly spherical, the diameter of the med... | 5. A.
Let the radius of the large melon be $R$, and the volumes of the large, medium, and small melons be $V_{\text {large }}, V_{\text {medium }}$, and $V_{\text {small }}$ respectively. Then, by the formula for the volume of a sphere, we have
$$
\begin{array}{l}
V_{\text {large }}: V_{\text {medium }}: V_{\text {sma... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 723,139 |
6. In the third year of a certain school, there are four classes, each class selects one male and one female table tennis player to form a mixed doubles team for the grade. Then, among the four mixed doubles pairs, the probability that no pair of players are from the same class is ( ).
(A) $\frac{5}{12}$
(B) $\frac{4}{... | 6. C.
4 men and 4 women forming four mixed doubles pairs have a total of 24 combinations.
Let the male and female players from classes one, two, three, and four be $A_{1}, B_{1}, A_{2}, B_{2}, A_{3}, B_{3}$, and $A_{4}, B_{4}$. Then, the scenarios where no pair of players are from the same class are as follows:
$$
\be... | C | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 723,140 |
4. When $n$ is a positive integer, it is defined that
$$
n!=n \times(n-1) \times \cdots \times 2 \times 1 \text {, }
$$
which is called the factorial of $n$ (for example, $10!=10 \times 9 \times \cdots \times 2 \times 1$ $=3628800$). Therefore, in 2010!, the total number of zeros at the end is $\qquad$ | 4.501.
The number of trailing zeros depends on the number of factors of 10. Since $10=2 \times 5$, and it is clear that in 2010! the number of factors of 2 is greater than the number of factors of 5, we only need to find the number of prime factors of 5 in 2010!, which gives us the number of trailing zeros.
Among the... | 501 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,144 |
Three, (18 points) Solve the system of equations:
$$
\left\{\begin{array}{l}
2(x+y+z)-5 \sqrt{x+y+z+5}=2, \\
\frac{x}{3}=\frac{y}{4}=\frac{z}{5} .
\end{array}\right.
$$ | Three, let $\sqrt{x+y+z+5}=t \geqslant 0$. Then $x+y+z=t^{2}-5$.
Substitute into equation (1) to get
$$
2\left(t^{2}-5\right)-5 t=2 \text {, }
$$
which simplifies to $2 t^{2}-5 t-12=0$.
Solving this, we get $t=4$ or $t=-\frac{3}{2}$ (discard).
From $\sqrt{x+y+z+5}=4 \Rightarrow x+y+z=11$.
From equation (2) and the pro... | x=\frac{11}{4}, y=\frac{11}{3}, z=\frac{55}{12} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,145 |
Four, (16 points) Function
$$
y=\left|x^{2}-1\right|+\left|2 x^{2}-1\right|+\left|3 x^{2}-1\right|
$$
Find the range of the independent variable $x$ when the function achieves its minimum value of 1. | Let $x^{2}=t(t \geqslant 0)$. Then
$$
y=|t-1|+|2 t-1|+|3 t-1| \text {. }
$$
When $0 \leqslant t < 1$,
$$
\begin{array}{l}
y=t-1+2 t-1+3 t-1=6 t-3 \\
\Rightarrow y>3 .
\end{array}
$$
In summary, when $\frac{1}{3} \leqslant t \leqslant \frac{1}{2}$, $y_{\text {min }}=1$.
Thus, from $\frac{1}{3} \leqslant x^{2} \leqslan... | -\frac{\sqrt{2}}{2} \leqslant x \leqslant-\frac{\sqrt{3}}{3} \text { or } \frac{\sqrt{3}}{3} \leqslant x \leqslant \frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,146 |
Five. (16 points) As shown in Figure 2, given that the circumcenter of $\triangle ABC$ is $O$, draw any circle passing through points $B$ and $C$, intersecting the extensions of $AB$ and $AC$ at points $E$ and $F$ respectively. Prove: $AO \perp EF$.
保留源文本的换行和格式，直接输出翻译结果如下：
Five. (16 points) As shown in Figure 2, give... | As shown in Figure 6, construct the circumcircle $\odot O$ of $\triangle A B C$, and extend $A O$ to intersect $\odot O$ and $E F$ at points $D$ and $G$ respectively. Connect $C D$. It is easy to see that
$$
\begin{array}{l}
\angle A E G=\angle A C B, \\
\angle B A D=\angle B C D, \angle A C D=90^{\circ} .
\end{array}
... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,147 |
Example 5 On a plane, there are seven points, and some line segments can be connected between them, so that any three points among the seven must have at least one line segment connecting a pair of them. How many line segments are needed at least? Prove your conclusion.
$(2002$, Shanghai Junior High School Mathematics ... | (1) If point $A$ is not an endpoint of any line segment, then every two of the other six points are connected by a line segment, totaling $\frac{1}{2} \times 6 \times 5=15$ line segments;
(2) If point $A$ is the endpoint of only one line segment, then every two of the five points not connected to $A$ are connected by a... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,148 |
Six. (16 points) As shown in Figure 3, there is a fixed point $P$ inside $\angle M A N$. It is known that $\tan \angle M A N=3, P$ is at a distance $P D=12$ from the line $A N$, and $A D=30$. A line through $P$ intersects $A N$ and $A M$ at points $B$ and $C$ respectively. Find the minimum area of $\triangle A B C$.
-... | Six, Solution 1
As shown in Figure 7, let $D B=$
$x$, draw $C E \perp A B$,
$P F / / C A$.
Let $A E=t$. Then
$$
\begin{array}{l}
\tan \angle M A N=3 \\
\Rightarrow C E=3 t .
\end{array}
$$
Similarly, by $P D=12 \Rightarrow F D=4$.
By $\triangle A B C \backsim \triangle F B P$, we get $\frac{A B}{C E}=\frac{F B}{P D}$,... | 624 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,149 |
1. The range of the function $f(x)=\sqrt{x-5}-\sqrt{24-3 x}$ is $\qquad$ . | $-1 \cdot[-3, \sqrt{3}]$.
It is known that the domain of $f(x)$ is $[5,8]$, and $f(x)$ is an increasing function on $[5,8]$.
Therefore, the range of $f(x)$ is $[-3, \sqrt{3}]$. | [-3, \sqrt{3}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,150 |
2. Given the function $y=\left(a \cos ^{2} x-3\right) \sin x$ has a minimum value of -3. Then the range of the real number $a$ is $\qquad$ | 2. $-\frac{3}{2} \leqslant a \leqslant 12$.
Let $\sin x=t$. Then, the original function becomes $g(t)=\left[a\left(1-t^{2}\right)-3\right] t$.
From the fact that the minimum value of $g(t)$ in $[-1,1]$ is -3, we get $a t\left(1-t^{2}\right)+3(1-t) \geqslant 0$,
which means $(1-t)[a t(1+t)+3] \geqslant 0$.
Thus, $a\lef... | -\frac{3}{2} \leqslant a \leqslant 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,151 |
3. The number of lattice points (points with integer coordinates) inside the region (excluding the boundary) bounded by the right branch of the hyperbola $x^{2}-y^{2}=1$ and the line $x=100$ is $\qquad$ | 3.9800 .
By symmetry, we only need to consider the situation above the $x$-axis first.
Let the line $y=k(k=1,2, \cdots, 99)$ intersect the right branch of the hyperbola at point $A_{k}$, and intersect the line $x=100$ at point $B_{k}$. Then the number of integer points inside the segment $A_{k} B_{k}$ is $99-k$. Ther... | 9800 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,152 |
4. Given that $\left\{a_{n}\right\}$ is an arithmetic sequence with a non-zero common difference, $\left\{b_{n}\right\}$ is a geometric sequence, where $a_{1}=3, b_{1}=1, a_{2}=b_{2}$, and for any number $n$, $a_{n}=\log _{\alpha} b_{n}+\beta$. Then $\alpha+\beta=$ $\qquad$ . | 4. $\sqrt[3]{3}+3$
Let $\left\{a_{n}\right\}$ have a common difference of $d$, and $\left\{b_{n}\right\}$ have a common ratio of $q$. Then
$$
\left\{\begin{array}{l}
3+d=q, \\
3(3+4 d)=q^{2} .
\end{array}\right.
$$
Solving, we get $d=6, q=9$.
Thus, $3+6(n-1)=\log _{\alpha} 9^{n-1}+\beta$ holds for all
positive intege... | \sqrt[3]{3}+3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,153 |
5. The function $f(x)=a^{2 x}+3 a^{x}-2(a>0, a \neq 1)$ has a maximum value of 8 on the interval $x \in[-1,1]$. Then its minimum value on this interval is $\qquad$ . | 5. $-\frac{1}{4}$.
Let $a^{x}=y$. Then the original function becomes
$$
g(y)=y^{2}+3 y-2 \text {, }
$$
and $g(y)$ is increasing on $\left(-\frac{3}{2},+\infty\right)$.
When $01$, $y \in\left[a^{-1}, a\right]$,
$$
g(y)_{\max }=a^{2}+3 a-2=8 \Rightarrow a=2 \text {. }
$$
Thus, $g(y)_{\min }=2^{-2}+3 \times 2^{-1}-2=-\... | -\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,154 |
6. Two people take turns throwing dice, each throwing two at a time. The first person to get a sum greater than 6 on the two dice wins; otherwise, the other person throws. What is the probability that the first person to throw wins? $\qquad$ | 6. $\frac{12}{17}$.
The probability of the sum of the points on two dice being greater than 6 is $\frac{21}{36}=\frac{7}{12}$, thus, the probability of the first player winning is
$$
\frac{7}{12}+\left(\frac{5}{12}\right)^{2} \times \frac{7}{12}+\left(\frac{5}{12}\right)^{4} \times \frac{7}{12}+\cdots=\frac{12}{17} .
... | \frac{12}{17} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,155 |
7. Given a regular triangular prism $A B C-A_{1} B_{1} C_{1}$ with all 9 edges of equal length, $P$ is the midpoint of edge $C C_{1}$, and the dihedral angle $B-A_{1} P-B_{1}=\alpha$. Then $\sin \alpha=$ $\qquad$ | 7. $\frac{\sqrt{10}}{4}$.
Solution 1 As shown in Figure 1, establish a spatial rectangular coordinate system with the line containing $AB$ as the $x$-axis, the midpoint $O$ of line segment $AB$ as the origin, and the line containing $OC$ as the $y$-axis.
Let the edge length of the regular triangular prism be 2. Then
... | \frac{\sqrt{10}}{4} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,156 |
9. (16 points) Given the function
$$
\begin{array}{l}
f(x)=a x^{3}+b x^{2}+c x+d(a \neq 0), \\
\text { when } 0 \leqslant x \leqslant 1 \text {, }|f^{\prime}(x)| \leqslant 1 .
\end{array}
$$
Try to find the maximum value of $a$. | II. Notice that $f^{\prime}(x)=3 a x^{2}+2 b x+c$. From $\left\{\begin{array}{l}f^{\prime}(0)=c, \\ f^{\prime}\left(\frac{1}{2}\right)=\frac{3}{4} a+b+c, \\ f^{\prime}(1)=3 a+2 b+c,\end{array}\right.$
we get $3 a=2 f^{\prime}(0)+2 f^{\prime}(1)-4 f^{\prime}\left(\frac{1}{2}\right)$.
Then $3|a|=\left|2 f^{\prime}(0)+2 f... | \frac{8}{3} | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 723,157 |
Example 6 There are $n$ different points on a plane, and a line segment is connected between every pair of points. The midpoint of each line segment is colored red.
(1) Prove: The number of red points on the plane is not less than $2 n-3$;
(2) Design a special case where the number of red points is equal to $2 n-3$.
(1... | (1) Among all the points, find the two points $M$ and $N$ that are farthest apart, and let their distance be $d$.
As shown in Figure 3, draw circles $C_{M}$ and $C_{N}$ with $M$ and $N$ as centers and $\frac{1}{2} d$ as the radius. Among the remaining $n-2$ points, for any point $P$, since $\frac{1}{2} M P \leqslant \f... | 2 n-3 | Geometry | proof | Yes | Yes | cn_contest | false | 723,158 |
10. (20 points) Given two moving points \( A\left(x_{1}, y_{1}\right) \) and \( B\left(x_{2}, y_{2}\right) \) on the parabola \( y^{2}=6 x \), where \( x_{1} \neq x_{2} \) and \( x_{1}+x_{2}=4 \), the perpendicular bisector of segment \( AB \) intersects the \( x \)-axis at point \( C \). Find the maximum value of the ... | 10. As shown in Figure 3, let the midpoint of line segment $AB$ be $M\left(x_{0}, y_{0}\right)$.
Then
$$
\begin{array}{l}
x_{0}=\frac{x_{1}+x_{2}}{2}=2 \text {, } \\
y_{0}=\frac{y_{1}+y_{2}}{2} \text {, } \\
k_{A B}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\
=\frac{y_{2}-y_{1}}{\frac{y_{2}^{2}}{6}-\frac{y_{1}^{2}}{6}}=\frac{6}... | \frac{14 \sqrt{7}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,159 |
11. (20 points) Prove: The equation $2 x^{3}+5 x-2=0$ has exactly one real root $r$, and there exists a unique strictly increasing sequence of positive integers $\left\{a_{n}\right\}$, such that
$$
\frac{2}{5}=r^{a_{1}}+r^{a_{2}}+\cdots \text {. }
$$ | 11. Let $f(x)=2 x^{3}+5 x-2$. Then
$$
f^{\prime}(x)=6 x^{2}+5>0 \text {. }
$$
Therefore, $f(x)$ is strictly increasing.
$$
\text { Also, } f(0)=-20 \text {, so } f(x)
$$
has a unique real root $r \in\left(0, \frac{1}{2}\right)$. Thus,
$$
\begin{array}{l}
2 r^{3}+5 r-2=0, \\
\frac{2}{5}=\frac{r}{1-r^{3}}=r+r^{4}+r^{7}... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,160 |
One. (40 points) Given an acute triangle $\triangle ABC$ with circumcenter $O$, $K$ is a point on side $BC$ (not the midpoint of side $BC$), and $D$ is a point on the extension of line segment $AK$. Line $BD$ intersects $AC$ at point $N$, and line $CD$ intersects $AB$ at point $M$. Prove that if $OK \perp MN$, then poi... | I. Proof by contradiction.
If $A, B,$
$D, C$ are not concyclic, as shown in Figure 4, let the circumcircle of $\triangle ABC$ intersect $AD$ at point $E$. Connect $BE$ and extend it to intersect line $AN$ at point $Q$. Connect $CE$ and extend it to intersect line $AM$ at point $P$. Connect $PQ$.
Notice that
$P K^{2}=$ ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,161 |
Sure, here is the translated text:
```
(40 points) Let $k$ be a given positive integer,
\[
\begin{array}{l}
r=k+\frac{1}{2} . \\
\text { Let } f^{(1)}(r)=f(r)=r\lceil r\rceil, \\
f^{(l)}(r)=f\left(f^{(l-1)}(r)\right)(l \geqslant 2) .
\end{array}
\]
Prove: There exists a positive integer $m$, such that $f^{(m)}(r)$ is... | Let $v_{2}(n)$ denote the exponent of 2 in the prime factorization of the positive integer $n$. Then, when $m=v_{2}(k)+1$, $f^{m}(r)$ is an integer.
We will use mathematical induction on $v_{2}(k)=v$.
When $v=0$, $k$ is odd, and $k+1$ is even. In this case,
$$
f(r)=\left(k+\frac{1}{2}\right)\left\lceil k+\frac{1}{2}\ri... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,162 |
Three. (50 points) Given an integer $n(n>2)$, let positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{k} \leqslant 1(k=1,2, \cdots, n)$, and let
$$
A_{k}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}(k=1,2, \cdots, n) .
$$
Prove: $\left|\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n} A_{k}\right|<\frac{n-1}{2}$. | Three, from $00$, we have
$$
|x-y|<\max \{x, y\} \text {. }
$$
Thus, for $1 \leqslant k \leqslant n-1$, we have
$$
\begin{array}{l}
\left|A_{n}-A_{k}\right|=\left|\frac{1}{n} \sum_{i=k+1}^{n} a_{i}-\left(\frac{1}{k}-\frac{1}{n}\right) \sum_{i=1}^{k} a_{i}\right| \\
<\max \left\{\frac{1}{n} \sum_{i=k+1}^{n} a_{i},\left... | proof | Inequalities | proof | Yes | Yes | cn_contest | false | 723,163 |
Four, (50 points) A password lock's password setting involves assigning one of the two numbers, 0 or 1, to each vertex of a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, and coloring each vertex one of two colors, red or blue, such that for any two adjacent vertices, at least one of the number or color is the s... | For this type of password lock setting, if the numbers assigned to two adjacent vertices are different, mark the edge between them with $a$; if the colors are different, mark it with $b$; if both the number and color are the same, mark it with $c$. Thus, for a given setting at point $A_{1}$ (with 4 possible settings), ... | 3^n + 1 \text{ if } n \text{ is odd, } 3^n + 3 \text{ if } n \text{ is even} | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,164 |
1. Let $a, b, c \in \{0,1, \cdots, 9\}$. If the quadratic equation $a x^{2}+b x+c=0$ has rational roots, prove: the three-digit number $abc$ is not a prime number.
（Zhang Pengcheng, problem contributor） | 1. Proof by contradiction.
Assume $\overline{a b c}=p$ is a prime number. Then, for the quadratic equation $f(x) = a x^{2} + b x + c = 0$, the rational roots are
$$
x_{1,2} = \frac{-b \pm \sqrt{b^{2} - 4 a c}}{2 a},
$$
It is easy to see that $b^{2} - 4 a c$ is a perfect square, and $x_{1}, x_{2}$ are both negative, a... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,165 |
2. For the set $A=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$, denote $P(A)=a_{1} a_{2} \cdots a_{m}$. Let $A_{1}, A_{2}, \cdots, A_{n}\left(n=\mathrm{C}_{2010}^{99}\right)$ be all the 99-element subsets of the set $\{1,2, \cdots, 2010\}$. Prove that: $2011 \mid \sum_{i=1}^{n} P\left(A_{i}\right)$.
(Ye Yongnan, problem... | 2. Proof 1 For each 99-element subset $A_{i}=\left\{a_{1}, a_{2}, \cdots, a_{99}\right\}$ of the set $\{1,2, \cdots, 2010\}$, it corresponds one-to-one with $B_{i}=\left\{b_{1}, b_{2}, \cdots, b_{99}\right\}$, where,
$$
b_{k}=2011-a_{k}(k=1,2, \cdots, 99) \text {. }
$$
Since $\sum_{k=1}^{99}\left(a_{k}+b_{k}\right)=99... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 723,166 |
3. As shown in Figure 1, given that the incircle $\odot I$ of $\triangle A B C$ touches sides $A B$ and $B C$ at points $F$ and $D$ respectively, and lines $A D$ and $C F$ intersect $\odot I$ at another point $H$ and $K$ respectively. Prove:
$$
\frac{F D \cdot H K}{F H \cdot D K}=3 \text {. }
$$
(Xiong Bin) | 3. Let $A F=x, B F=y, C D=z$.
By Stewart's Theorem, we have
$A D^{2}=\frac{B D}{B C} \cdot A C^{2}+\frac{C D}{B C} \cdot A B^{2}-B D \cdot D C$ $=\frac{y(x+z)^{2}+z(x+y)^{2}}{y+z}-y z=x^{2}+\frac{4 x y z}{y+z}$.
By the Power of a Point Theorem, we get $A H=\frac{A F^{2}}{A D}=\frac{x^{2}}{A D}$.
Thus, $H D=A D-A H=\fr... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,167 |
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